STD 9 CBSE MATHS CH 1 SVADHYAY 1.6

Q 1 Ex-3.5 | STD 6 MATHS | Chap-3: PLAYING WITH NUMBERS| PART-21 | NCERT |

Q 6 Ex-3.5 | STD 6 MATHS | Chap-3: PLAYING WITH NUMBERS| PART-25 | NCERT |

Q 4 Ex-3.5 | STD 6 MATHS | Chap-3: PLAYING WITH NUMBERS| PART-23 | NCERT |

Q 5 Ex-3.5 | STD 6 MATHS | Chap-3: PLAYING WITH NUMBERS| PART-24 | NCERT |

Q 7 Ex-3.5 | STD 6 MATHS | Chap-3: PLAYING WITH NUMBERS| PART-26 | NCERT |

Q 2, 3 Ex-3.5 | STD 6 MATHS | Chap-3: PLAYING WITH NUMBERS| PART-22 | NCERT |

Class 10 Exercise 10.2 Question 1 | CBSE Class 10 Ex -10.2 Q1 | NCERT | Class 10 Maths Chapter 10

If A={1,2,3},\ B="{"1,4,6,9"]"a n d\ R is a relation from A to B defined by x is greater than y . The range of R is A. {1,4,6,9} B. {4,6,9} C. {1} D. none of these

Show that 1 + 1/3 + (1.3)/(3.6) + (1.3.5)/(3.6.9) + ….. = sqrt3

The total number of electrons in 1.6 g of CH_(4) to that in 1.8 g of H_(2)O